A 60 p F capacitor is fully charged by a 20 v supply.
Question:

A $60 p F$ capacitor is fully charged by a $20 \vee$ supply. It is then disconnected from the supply and is connected to another uncharged $60 p F$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $\mathrm{nJ}$ )

Solution:

(6) In the first condition, electrostatic energy is

$U_{i}=\frac{1}{2} C V_{0}^{2}=\frac{1}{2} \times 60 \times 10^{-12} \times 400=12 \times 10^{-9} \mathrm{~J}$

In the second condition $U_{F}=\frac{1}{2} C^{\prime} V^{\prime 2}$

$U_{f}=\frac{1}{2} 2 C \cdot\left(\frac{V_{0}}{2}\right)^{2} \quad\left(\because C^{\prime}=2 C, V^{\prime}=\frac{V_{0}}{2}\right)$

$=\frac{1}{4} \times 60 \times 10^{-12} \times(20)^{2}=6 \times 10^{-9} \mathrm{~J}$

Energy lost $=U_{i}-U_{f}=12 \times 10^{-9} J-6 \times 10^{-9} J=6 \mathrm{~nJ}$

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